WLOG $a \geq b \geq c$ so $1 \geq 2c-1<2a$ hence $a=2c-1$. From here we get: $$b \vert 2a-1=4c-3$$$$c \vert 2b-1$$Write $b=\frac{kc+1}{2}$ for some $k$ then: $$a \geq b \geq c \Rightarrow 1 \geq c(2-k) \quad \quad c(4-k) \geq 3$$If $k=1$ then the only solution is $c=1$. For $k \geq 4$ the second inequality gives a contradiction. So we just need to check $k=2,3$. $k=2$: This gives a contradiction as $b$ is an integer. $k=3$: We need $3c+1 \vert 26 \Rightarrow c=4$ but this doesn't work for $b$ Hence we must have $a=b=c=1$ As @below points out we also need to consider the case $a \leq b \leq c$

sbealing wrote: WLOG $a \geq b \geq c$ so $1 \geq 2c-1<2a$ hence $a=2c-1$. From here we get: $$b \vert 2a-1=4c-3$$$$c \vert 2b-1$$Write $b=\frac{kc+1}{2}$ for some $k$ then: $$a \geq b \geq c \Rightarrow 1 \geq c(2-k) \quad \quad c(4-k) \geq 3$$If $k=1$ then the only solution is $c=1$. For $k \geq 4$ the second inequality gives a contradiction. So we just need to check $k=2,3$. $k=2$: This gives a contradiction as $b$ is an integer. $k=3$: We need $3c+1 \vert 26 \Rightarrow c=4$ but this doesn't work for $b$ Hence we must have $a=b=c=1$ Why if $k=1$ then $c=1$? Also we can`t suppose that $a \geq b \geq c$ because condition is cyclic, not symmetric. I think 7, 13, 25 works.

Here we go. Without loss of generality, let $a$ be the smallest amongst. Note that, from $a\mid 2c-1$, we must have $a$ is odd (similarly, $b$ and $c$). If $a=1$, we get $(1,1,1)$. If $a=3$, then $b\mid 5$, and since $b>a$ we have that $b=5$. Thus, $c\mid 9$, hence, $c\in\{3,9\}$, both of which are impossible. Handle $a=5,7,9,11$ like this, and suppose that $a\geq 13$. The given condition implies, $abc\mid (2a-1)(2b-1)(2c-1)$, namely, $$ 8abc-4ab-4ac-4bc+2a+2b+2c-1 \equiv 0\pmod{abc}\implies abc \mid 4ab+4ac+4bc-2a-2b-2c+1. $$This yields, $$ abc\leq 4(ab+bc+ca)-2(a+b+c)+1 < 4(ab+bc+ca)\implies \frac{1}{a}+\frac{1}{b}+\frac{1}{c}>\frac{1}{4}. $$On the other hand, since $a$ is the smallest, and $a\geq 13$, we have, $1/a+1/b+1/c<3/13$, which is a contradiction.

We will show that the only solutions are cyclic permutations of $(7,13,25)$, the solution that richrow gave. Write $2a-1=kb,2b-1=lc,2c-1=ma$ then $k,l,m\ge 1, klm\equiv 1(2)$ and $kb<2a,lc<2b,ma<2b$ hence $klm(abc)<8abc$ and $klm<8$. We will sharpen this a bit now. We have $a={{kb+1}\over 2},c={{2b-1}\over l}$ and putting this into the other equation gives, after some rewriting $(8-klm)b=lm+2l+4$. Cyclic permutation gives $$(8-klm)b=lm+2l+4,(8-klm)a=kl+2k+4,(8-klm)c=mk+2m+4$$If $klm=7$ then we may assume $k=7,l=1,m=5$ and we find $a=25,b=7,c=13$. If $klm=5$ then $k=5,l=1,m=1$ and $3b=7$, not possible. If $klm=3$ then $k=3,l=1,m=1$ and $5b=7$, not possible. If $klm=1$ then $7b=7$ etc, hence $a=b=c=1$, not possible.

alexheinis wrote: We will show that the only solutions are cyclic permutations of $(7,13,25)$, the solution that richrow gave. Write $2a-1=kb,2b-1=lc,2c-1=ma$ then $k,l,m\ge 1, klm\equiv 1(2)$ and $kb<2a,lc<2b,ma<2b$ hence $klm(abc)<8abc$ and $klm<8$. We will sharpen this a bit now. We have $a={{kb+1}\over 2},c={{2b-1}\over l}$ and putting this into the other equation gives, after some rewriting $(8-klm)b=lm+2l+4$. Cyclic permutation gives $$(8-klm)b=lm+2l+4,(8-klm)a=kl+2k+4,(8-klm)c=mk+2m+4$$If $klm=7$ then we may assume $k=7,l=1,m=5$ and we find $a=25,b=7,c=13$. If $klm=5$ then $k=5,l=1,m=1$ and $3b=7$, not possible. If $klm=3$ then $k=3,l=1,m=1$ and $5b=7$, not possible. If $klm=1$ then $7b=7$ etc, hence $a=b=c=1$, not possible. Nice solution from alexheinis.